Silver nitrate is an ionic bond because it is made up of metal, sliver, and a non-metal, nitrogen and oxygen. It is also a polyatomic ion (you only find polyatomic ions in ionic bonds).
<span>1 drop is approximately 0.05mL. Since 0.500L of 0.550M NH4Cl contains 0.275mol of substance (calculated by using c=n/V formula), equal amount of substance of NH3 is needed to neutralize this solution (since pH of 7 is neutral solution). Thus, we need 0.0275L of NH3, i.e. around 550 drops.</span>
Answer:
146 g/mol → option b.
Explanation:
This is a problem about the freezing point depression. The formula for this colligative property is:
ΔT = Kf . m . i
We assume i = 1, so our compound is not electrolytic.
ΔT = Freezing T° of pure solvent - Freezing T° of solution = 1.02 °C
m = molality (mol of solute/kg of solvent)
We convert the grams of solvent (benzene) to kg → 250 g . 1 kg/1000 = 0.250 kg.
We replace → 1.02°C = 5.12°C/mol/kg . mol/ 0.250kg . 1
1.02°C / 5.12 mol/kg/°C = mol/ 0.250kg
0.19922 mol/kg = mol/ 0.250kg
mol = 0.19922 . 0.250kg → 0.0498 mol
molar mass = g/mol → 7.27 g / 0.0498mol = 146 g/mol